In Delta ABC,m angle B = 90^{circ} and overli | vedclass

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(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other. By the geometr

Vedclass · Accepted Answer

$AB = 8, BC = 8\sqrt{3}$

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(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other. By the geometric mean theorem,$BM^2 = AM \cdot CM$. Given $AM = 4$ and $CM = 12$,$BM^2 = 4 \cdot 12 = 48$,so $BM = \sqrt{48} = 4\sqrt{3}$. In $\Delta ABM$,by the Pythagorean theorem,$AB^2 = AM^2 + BM^2 = 4^2 + (4\sqrt{3})^2 = 16 + 48 = 64$. Thus,$AB = \sqrt{64} = 8$. In $\Delta CBM$,by the Pythagorean theorem,$BC^2 = CM^2 + BM^2 = 12^2 + (4\sqrt{3})^2 = 144 + 48 = 192$. Thus,$BC = \sqrt{192} = 8\sqrt{3}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 12$,find $AB$ and $BC$.

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